## Exponential and Logarithmic Equations |

**Contents:** This page corresponds to **§** **4.4
(p. 348)** of the text.

Suggested problems from text:

p. 355 #13,17,21,25,27,29,31,51,53,57,59,81,89

## Exponential Equations

## Logarithmic Equations

Some exponential equations can be solved by using the fact that exponential functions are one-to-one. In other words, an exponential function does not take two different values to the same number.

Example 1.

3

^{x}= 93

^{x}= 3^{2}The function f(x) = 3

^{x}is one-to-one, so it does not take two different values to 9, so x must equal 2.x = 2

The equation in example 1 was easy to solve because we could express 9 as a power of 3. However, it is often necessary to use a logarithm when solving an exponential equation.

Example 2.

e

^{x}= 20We are going to use the fact that the natural logarithm is the inverse of the exponential function, so ln e

^{x}= x, by logarithmic identity 1. We must take the natural logarithm of both sides of the equation.ln e

^{x}= ln 20Now the left hand side simplifies to x, and the right hand side is a number. It is approximately 2.9957.

x = 2.9957

Use a calculator to check the answer we found to the equation in example 2.

Example 3.

5

^{x}= 16 We will solve this equation in two different ways.First Approach: We use the fact that log

_{5}5^{x}= x (logarithmic identity 1 again).5

^{x}= 16log

_{5}5^{x}= log_{5}16x = log

_{5}16x = ln 16 / ln 5, by the change-of-base formula.

x = 1.7227 (approximately)

Second Approach: We will use the natural logarithm and property 3.

5

^{x}= 16 Take the natural logarithm of both sides.ln 5

^{x}= ln 16x ln 5 = ln 16

x = ln 16 / ln 5

x = 1.7227 (approximately)

We could have used any logarithm with the second approach. The second approach is the one that you see most often.

Use a calculator to check the answer we found to the equation in example 3.

Equations like that in the next example occur frequently in applications.

Example 4.

200 e

^{0.07t}= 500We first isolate the exponential part by dividing both sides of the equation by 200.

e

^{0.07t}= 2.5Now we take the natural logarithm of both sides.

ln e

^{0.07t}= ln 2.5The left hand side simplifies to 0.07t, by logarithmic identity 1.

0.07t = ln 2.5

t = ln (2.5) / 0.07

t = 13.1 (approximately)

Solve the following equations and check the answers.

(a) 3

^{x}= 10(b) 150 e

^{0.05 t}= 350

When solving exponential equations we frequently used logarithmic identity 1 because it involves applying a logarithmic function to "undo" the effect of an exponential function. When dealing with logarithmic equations we will use logarithmic identity 2 where an exponential function is applied to "undo" the effect of a logarithmic function.

Example 5.

2 log x = 12

We want to isolate the log x, so we divide both sides by 2.

log x = 6

Since log is the logarithm base 10, we apply the exponential function base 10 to both sides of the equation.

10

^{log x}= 10^{6}By logarithmic identity 2, the left hand side simplifies to x.

x = 10

^{6}= 1000000

Example 6.

7 + 3 ln x = 15 First isolate ln x.

3 ln x = 8

ln x = 8/3

Now apply the exponential function to both sides.

e

^{ln x}= e^{8/3}x = e

^{8/3}This is the

exactanswer. If you use a calculator to evaluate this expression, you will have anapproximationto the answer.x is approximately equal to 14.39.

Check the answers found in examples 5 and 6.

Example 7.

ln (x + 4) + ln (x - 2) = ln 7

First we use property 1 of logarithms to combine the terms on the left.

ln (x + 4)(x - 2) = ln 7

Now apply the exponential function to both sides.

e

^{ln (x + 4)(x - 2)}= e^{ln 7}The logarithmic identity 2 allows us to simplify both sides.

(x + 4)(x - 2) = 7

x

^{2}+ 2x - 8 = 7x

^{2}+ 2x - 15 = 0(x - 3)(x + 5) = 0

x = 3 or x = -5

x = 3 checks, for ln 7 + ln 1 = ln 7.

x = -5

does notcheck, for when we try to substitute -5 for x in the original equation we are taking the natural logarithm ofnegativenumbers, which is not defined.So, x = 3 is the only solution.